How do you evaluate $ {{\log }_{7}}49 $ ? \[\]

Hint: We recall the definition of logarithm and the identity involving power $ m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}} $ and the identity where base and argument are equal $ {{\log }_{b}}b=1 $ . We express 49 as $ 49={{7}^{2}} $ and use these identities to simplify. \[\]

Hint:
We know that the logarithm is the inverse operation to exponentiation. That means the logarithm of a given number $ x $ is the exponent to which another fixed number, the base $ b $ must be raised, to produce that number $ x $, which means if $ {{b}^{y}}=x $ then the logarithm denoted as log and calculated as
$\Rightarrow$ \[{{\log }_{b}}x=y\]
Here the base is always positive and is never equal to 1 $ \left( b > 0,b\ne 1 \right) $ . Here $ x $ is called argument of the logarithm and is always positive.
We know that when base and argument of a logarithm are equal we have;
$\Rightarrow$ \[{{\log }_{b}}b=1\]
We know the logarithmic identity involving power $ m\ne 0 $ where $ m $ is real number as
$\Rightarrow$ \[m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}}\]
We are asked to evaluate the value of $ {{\log }_{7}}49 $ . Here we have the base as $ b=7 $ and argument of the logarithm is $ x=49 $ . We know from exponentiation that we can write 49 as $ 49=7\times 7={{7}^{2}} $ . So we have
$\Rightarrow$ \[{{\log }_{7}}49={{\log }_{7}}{{7}^{2}}\]
We see in the right hand side of above step that logarithm is in the form of $ {{\log }_{b}}{{x}^{m}} $ where the power is $ m=7 $ . We use the identity $ m{{\log }_{b}}x={{\log }_{b}}{{x}^{m}} $ in the above step to have;
\[\Rightarrow {{\log }_{7}}49=2{{\log }_{7}}7\]
We see in the right hand side of above step that the argument and base of the logarithm is equal. So we use the identity $ {{\log }_{b}}b=1 $ to have;
\[\Rightarrow {{\log }_{7}}49=2\times 1=2\]
$\Rightarrow$ So the evaluated result is 2. \[\]

Note:
 We note that if we get large a number which is equal to large power of 7 for example $ {{7}^{5}} $ then we should use the prime factorization using divisibility rules. We know that a number is divisible by 7 when alternating sum and difference of numbers made from a triplet of digits from the right-hand side of the number is divisible by 7, for example, 368,851 is divisible by 7 because $ 851-369=483 $ is divisible by 7.